Taylor-Series at point x = 0

Question

We have $$ f(x) = (1+x)^{1/3} $$

I have to find the taylor series of f at the point x =0. The problem I am facing is that I do not know how often do I have to derive f

$$ f'(x) = \frac{1}{3}(1+x)^{-2/3} $$ $$ f''(x) = -\frac{2}{9}(1+x)^{-5/3} $$ $$ f'''(x) = \frac{10}{27}(1+x)^{-8/3} $$ $$ f^{4}(x) = -\frac{80}{81}(1+x)^{-11/3}$$

As you can see, I can derive infinity times. As you also can see, there is a certain pattern when you oberseve the derivatives. The pattern I recognised is something like that:

$$ f^{(n)}(x) = (-1)^{n+1}\frac{..}{3^n}(1+x)^{-\frac{3n-1}{3}} $$

I don't know what to put in ".." in the first fraction.

I also don't know if I am completely wrong or right. Thank you in anticipation.

Answer 1

Look at the following table, where the first column is the order of the derivative, and the second column the numerator of the fraction: $$\begin{align} 1&\quad-1\\ 2&\quad1\cdot2\\ 3&\quad1\cdot2\cdot5\\ 4&\quad1\cdot2\cdot5\cdot8\\ 5&\quad1\cdot2\cdot5\cdot8\cdot11 \end{align}$$ Can you see the pattern?

Answer 2

From \begin{align*} f'(x) &= \frac{1}{3}(1+x)^{-\color{blue}{2}/3}\\ f''(x) &= -\frac{2}{9}(1+x)^{-\color{blue}{5}/3} \\ f'''(x) &= \frac{10}{27}(1+x)^{-\color{blue}{8}/3} \\ f^{4}(x) &= -\frac{80}{81}(1+x)^{-11/3}\\ \end{align*} we conclude the numerator is \begin{align*} \color{blue}{2}\cdot \color{blue}{5}\cdot \color{blue}{8}=\prod_{j=1}^3 (3j-1)=8!!! \end{align*} with $n!!!=n\cdot (n-3)\cdot (n-6)\cdots $ the triple factorial.

In general $n$ we obtain \begin{align*} f^{(n)}(x) &= (-1)^{n+1}\frac{\prod_{j=1}^n (3j-1)}{3^n} (1+x)^{-\frac{3n-1}{3}} \\ &= (-1)^{n+1}\frac{(3n-1)!!!}{3^n} (1+x)^{-\frac{3n-1}{3}} \end{align*}

Answer 3

$\prod\limits_{i=0}^{i=n-2}(2+3i) = 3^{n-1}\prod\limits_{i=0}^{i=n-2}(i+\frac{2}{3}) = 3^{n-1}(n-2+\frac{2}{3})!=3^{n-1}(n-\frac{4}{3})!$